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alisha [4.7K]
2 years ago
14

What is the value of the element at b23 ?

Mathematics
1 answer:
Leya [2.2K]2 years ago
4 0

\qquad\qquad \textit{columns}\\ b=\begin{bmatrix} 6&-5&16\\ 3&22&\boxed{24}\\ 7&22&14 \end{bmatrix} \begin{array}{llll} r\\o\\w\\s \end{array}\qquad \impliedby \qquad \begin{array}{llll} b23\\ \textit{matrix "b"}\\ \textit{row 2}\\ \textit{column 3} \end{array}

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A survey determined that 5/19 of the seventh grade students at Mills Middle School preferred round chocolate candies to hard can
Kaylis [27]

Answer:3/19

Step-by-step explanation:

5/19 - 2/19 = 3/19

5 0
3 years ago
Suppose A and B are independent events if P(A) = 0.4 And P(B) = 0.1, what is P(A'uB)? APEX
Aloiza [94]
For Independent Events, P(A) × P(B) = P(A∩B)

so we have, P(A∩B) = 0.4×0.1 = 0.04

P(A') = 1 - 0.4 = 0.6

This information can be represented on a Venn diagram as shown below

P(A'∪B) means the union of everything that is not A with everything that is B

P(A'∪B) = 0.06 + 0.54 + 0.04 = 0.64


5 0
3 years ago
The answer is not 1465.5. Every month, Ms. Thomas pays her car loan through automatic payments (withdrawals) from her savings ac
Andrew [12]

Answer:

$244.25

Step-by-step explanation:

First, the question is asking what the change in her savings account balance is each month. So, you know that you are looking for the amount that is lost each month. When it says that at the end of the year, she has payed $2,931 , what is implied is that she has spent that much over 12 months. So all that is needed to do is divide her balance change over the year by 12 to find the monthly balance change.

4 0
3 years ago
d.) The expression that describes the network of logic gates: Is [A AND (NOT B)] OR (NOT B). Complete the input-output table for
marusya05 [52]
<span>Let R=A AND (NOT B)] OR (NOT B)
 If A=T, B=T
then ~B=F,
A and ~B = F
So R=F
 If A=T,
B=F ~B=T
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3 0
3 years ago
Lim<br> x-&gt;infinity (1+1/n)
FrozenT [24]

Answer:

^{ \lim}_{n \to \infty} (1+\frac{1}{n})=1

Step-by-step explanation:

We want to evaluate the following limit.


^{ \lim}_{n \to \infty} (1+\frac{1}{n})


We need to recall that, limit of a sum is the sum of the limit.


So we need to find each individual limit and add them up.

^{ \lim}_{n \to \infty} (1+\frac{1}{n})=^{ \lim}_{n \to \infty} (1) +^{ \lim}_{n \to \infty} \frac{1}{n}


Recall that, as n\rightarrow \infty,\frac{1}{n} \rightarrow 0 and the limit of a constant, gives the same constant value.



This implies that,


^{ \lim}_{n \to \infty} (1+\frac{1}{n})= 1 +0


This gives us,

^{ \lim}_{n \to \infty} (1+\frac{1}{n})= 1


The correct answer is D



5 0
3 years ago
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